Question

A pile driver hammer of mass 140 kg falls freely through a distance of 4.5 m to strike a pile of mass 390 kg and drives it 70 mm into the ground. The hammer does not rebound when driving the pile. Determine the average resistance of the ground.

Answer 1

The potential energy of the pile-driver is converted into kinetic energy. Thus

$$\text{potential energy} = \text{kinetic energy,}$$ $$mgH=\dfrac{mv^2}{2}$$

Find velocity of pile immediately after impact using principle of conservation of momentum

$$mv=(M+m)u$$

$$u=\dfrac{m}{M+m}v$$

The pile-driver and pile together have a mass $(m+M)$ and possess kinetic energy $\dfrac{(M+m)u^2}{2}.$

The change in potential energy of the driver as it moves in the pile and the pile as it moves through the ground

$$(M+m)g(0-h)$$

Let $R=$ the average resistance of the ground. Then

$$\text{work done}=R\times h$$

By the principle of conservation of energy

$$\dfrac{(M+m)u^2}{2}=(M+m)g(0-h)+R\times h$$

$$R=(M+m)g+\dfrac{(M+m)u^2}{2h}$$

$$R=(M+m)g+\dfrac{m^2v^2}{2h(M+m)}$$

$$R=(M+m)g+\dfrac{m^2gH}{h(M+m)}$$

$$(140kg+390kg)(9.81m/s^2)+\dfrac{(140kg)^2(9.81m/s^2)(4.5m)}{0.07m(140kg+390kg)}$$

$$R=28521N$$

The average resistance of the ground is 28.521 kN.

We see that the kinetic energy of the system at impact is not conserved.